I know that Joint Probability density function for two random functions $X$ and $Y$
$$P(XY) = P(X)\cdot P(Y|X)\tag{1}$$
But I just read in a set of lecture notes that for E(X)=E(Y)=0
$$E(XY) = E(X)\cdot E(Y|X)\tag{2}$$
I also know that $E(X) =$ Multiplication of $P(X)$ with $x$ and subsequent Integration over $x$
Hence, I don't see how equation (2) stands true. Any ideas? Is it wrongly written.
Your equation (1) is misleading. The joint density function of two random variables is not related to their product. Further we more commonly use $f$ for probability density functions, rather than $\mathsf P$. So the joint PDF is:
$$f_{X,Y}(x,y) = f_X(x)f_{Y\mid X}(y\mid x)$$
The expected value of the product of two random variables is then:
$$\begin{align} \mathsf E[XY] & = \iint_{\mathcal{X\times Y}} xy\; f_{X,Y}(x,y)\operatorname d y \operatorname d x \\[1ex] & = \iint_{\mathcal{X\times Y}} xy\;f_{Y\mid X}(y\mid x)f_X(x)\operatorname d y \operatorname d x \\[1ex] & = \int_\mathcal X x\left(\int_\mathcal{Y\mid X=x} y\;f_{Y\mid X}(y\mid x)\operatorname d y\right)f_X(x)\operatorname dx \\[1ex] & = \int_\mathcal X x\;\mathsf E[Y\mid X=x]\;f_X(x)\operatorname dx \\[1ex] & = \mathsf E[X\;\mathsf E[Y\mid X]] \end{align}$$
Which is not the same thing as: $\mathsf E[X]\;\mathsf E[Y\mid X]$